Optimal. Leaf size=134 \[ \frac {c^2 2^{-\frac {m}{2}+\frac {n}{2}+2} (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^n (1-\sin (e+f x))^{\frac {m-n}{2}} (g \cos (e+f x))^{-m-n} \, _2F_1\left (\frac {1}{2} (m-n-2),\frac {m-n}{2};\frac {1}{2} (m-n+2);\frac {1}{2} (\sin (e+f x)+1)\right )}{f g (m-n)} \]
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Rubi [A] time = 0.37, antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.089, Rules used = {2853, 2689, 70, 69} \[ \frac {c^2 2^{-\frac {m}{2}+\frac {n}{2}+2} (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^n (1-\sin (e+f x))^{\frac {m-n}{2}} (g \cos (e+f x))^{-m-n} \, _2F_1\left (\frac {1}{2} (m-n-2),\frac {m-n}{2};\frac {1}{2} (m-n+2);\frac {1}{2} (\sin (e+f x)+1)\right )}{f g (m-n)} \]
Antiderivative was successfully verified.
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Rule 69
Rule 70
Rule 2689
Rule 2853
Rubi steps
\begin {align*} \int (g \cos (e+f x))^{-1-m-n} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{2+n} \, dx &=\left ((g \cos (e+f x))^{-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^m\right ) \int (g \cos (e+f x))^{-1+m-n} (c-c \sin (e+f x))^{2-m+n} \, dx\\ &=\frac {\left (c^2 (g \cos (e+f x))^{-m-n} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{m+\frac {1}{2} (-m+n)} (c+c \sin (e+f x))^{\frac {1}{2} (-m+n)}\right ) \operatorname {Subst}\left (\int (c-c x)^{2-m+\frac {1}{2} (-2+m-n)+n} (c+c x)^{\frac {1}{2} (-2+m-n)} \, dx,x,\sin (e+f x)\right )}{f g}\\ &=\frac {\left (2^{1-\frac {m}{2}+\frac {n}{2}} c^3 (g \cos (e+f x))^{-m-n} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{\frac {m}{2}+\frac {n}{2}+\frac {1}{2} (-m+n)} \left (\frac {c-c \sin (e+f x)}{c}\right )^{\frac {m}{2}-\frac {n}{2}} (c+c \sin (e+f x))^{\frac {1}{2} (-m+n)}\right ) \operatorname {Subst}\left (\int \left (\frac {1}{2}-\frac {x}{2}\right )^{2-m+\frac {1}{2} (-2+m-n)+n} (c+c x)^{\frac {1}{2} (-2+m-n)} \, dx,x,\sin (e+f x)\right )}{f g}\\ &=\frac {2^{2-\frac {m}{2}+\frac {n}{2}} c^2 (g \cos (e+f x))^{-m-n} \, _2F_1\left (\frac {1}{2} (-2+m-n),\frac {m-n}{2};\frac {1}{2} (2+m-n);\frac {1}{2} (1+\sin (e+f x))\right ) (1-\sin (e+f x))^{\frac {m-n}{2}} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n}{f g (m-n)}\\ \end {align*}
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Mathematica [F] time = 142.84, size = 0, normalized size = 0.00 \[ \int (g \cos (e+f x))^{-1-m-n} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{2+n} \, dx \]
Verification is Not applicable to the result.
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fricas [F] time = 0.51, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\left (g \cos \left (f x + e\right )\right )^{-m - n - 1} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{n + 2}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (g \cos \left (f x + e\right )\right )^{-m - n - 1} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{n + 2}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 1.47, size = 0, normalized size = 0.00 \[ \int \left (g \cos \left (f x +e \right )\right )^{-1-m -n} \left (a +a \sin \left (f x +e \right )\right )^{m} \left (c -c \sin \left (f x +e \right )\right )^{2+n}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (g \cos \left (f x + e\right )\right )^{-m - n - 1} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{n + 2}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{n+2}}{{\left (g\,\cos \left (e+f\,x\right )\right )}^{m+n+1}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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